Assuming `H_(2)SO_(4)` to be completely ionised the `pH` of a `0.05M` aqueous of sulphuric acid is approximately

Options

A. `0.01`
B. `0.005`
C. `2`
D. `1`

Explanation

To find the pH of a 0.05 M solution of sulfuric acid (H₂SO₄) assuming complete ionization, we can follow these steps:

### Step 1: Understand the Ionization of Sulfuric Acid
Sulfuric acid (H₂SO₄) is a strong acid and is dibasic, meaning it can donate two protons (H⁺ ions) per molecule. The complete ionization can be represented as:
\[ \text{H}_2\text{SO}_4 \rightarrow 2 \text{H}^+ + \text{SO}_4^{2-} \]

### Step 2: Calculate the Concentration of H⁺ Ions
Since H₂SO₄ completely ionizes, for every 1 mole of H₂SO₄, we get 2 moles of H⁺ ions. Therefore, for a 0.05 M solution of H₂SO₄:
\[ \text{[H}^+\text{]} = 0.05 \, \text{M} \times 2 = 0.1 \, \text{M} \]

### Step 3: Use the pH Formula
The pH is calculated using the formula:
\[ \text{pH} = -\log[\text{H}^+] \]
Substituting the concentration of H⁺ ions:
\[ \text{pH} = -\log(0.1) \]

### Step 4: Calculate the Logarithm
The logarithm of 0.1 can be calculated as:
\[ \log(0.1) = -1 \]
Thus,
\[ \text{pH} = -(-1) = 1 \]

### Final Answer
The pH of a 0.05 M aqueous solution of sulfuric acid is approximately **1**.

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